a) $\Bigg(\dfrac{1-a\sqrt[]{a}}{1-\sqrt[]{a}}+\sqrt[]{a}\Bigg)\Bigg(\dfrac{1-\sqrt[]{a}}{1-a}\Bigg)^2$
$=\Bigg[\dfrac{(1-\sqrt[]{a})(a+\sqrt[]{a}+1)}{1-\sqrt[]{a}}+\sqrt[]{a}\Bigg].\Bigg[\dfrac{1-\sqrt[]{a}}{(1-\sqrt[]{a})(1+\sqrt[]{a})}\Bigg]^2$
$=(a+2\sqrt[]{a}+1).\dfrac{1}{(\sqrt[]{a}+1)^2}$
$=\dfrac{(\sqrt[]{a}+1)^2}{(\sqrt[]{a}+1)^2}$
$=1$ (điều phải chứng minh)
b) $\dfrac{a+b}{b^2}\sqrt[]{\dfrac{a^2b^4}{a^2+2ab+b^2}}$
$=\dfrac{a+b}{b^2}\sqrt[]{\dfrac{(ab^2)^2}{(a+b)^2}}$
$=\dfrac{a+b}{b^2}.\dfrac{|a|.b^2}{|a+b|}$
Vì $\left\{ \begin{array}{l}a+b>0\\b\neq 0\end{array} \right.$ nên:
$\dfrac{a+b}{b^2}.\dfrac{|a|.b^2}{|a+b|}$
$=\dfrac{a+b}{b^2}.\dfrac{|a|.b^2}{a+b}$
$=|a|$ (điều phải chứng minh)
